IODS course project
Paavo Laitinen
Introduction to Open Data Science - Course Project
About the project
Write a short description about the course and add a link to your GitHub repository here. This is an R Markdown (.Rmd) file so you should use R Markdown syntax.
# This is a so-called "R chunk" where you can write R code.
date()## [1] "Sun Nov 27 14:13:34 2022"The text continues here.
I hope everything is now installed correctly. Learning Git/Github seems useful and it is something that should prove to be useful in the future. I am a bit nervous about using R as I’m not very familar with it but we’ll see how this goes.
Thoughts about Exercise set 1
R really seems like a good tool for building graphs and plots and the logic seems pretty straightforward. However, it will take time to learn the syntax and all the commands. It is inevitable that learning new software feels at first like an avalanche of information, but this tutorial for R was nice in that it teaches one thing at a time.
Chapter 2) Regression and model validation
Describe the work you have done this week and summarize your learning.
- Describe your work and results clearly.
- Assume the reader has an introductory course level understanding of writing and reading R code as well as statistical methods.
- Assume the reader has no previous knowledge of your data or the more advanced methods you are using.
date()## [1] "Sun Nov 27 14:13:34 2022"1) Reading data
lrn14 <- read.table("learning2014.csv", sep=",", header=TRUE)
# dimensions of the data
dim(lrn14)## [1] 166 7# structure of the data
str(lrn14)## 'data.frame': 166 obs. of 7 variables:
## $ gender : chr "F" "M" "F" "M" ...
## $ age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: int 37 31 25 35 37 38 35 29 38 21 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ points : int 25 12 24 10 22 21 21 31 24 26 ...This is survey data from 2014 of Approaches to learning.This is a subset of original learning2014 data, from which 7 variables were picked:gender, age, attitude, deep, stra, surf and points. ‘Deep’(deep learning), ‘stra’ (strategic learning) and ‘surf’ (superficial learning) have been combined from their related items. You can find more information about the dataset from learning2014
2) Overview of data
library(ggplot2)
library(GGally)## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2p <- ggpairs(lrn14, mapping = aes(col = gender, alpha = 0.3), lower = list(combo = wrap("facethist", bins = 20)))
p
Looking at the data, there are more females than males. Age is the only variable that is clearly skewed (to the left), while other variables are more evenly distributed. Srongest correlations are found between deep learning and surface learning (-0.32), attitude and point (0.44). Additionally, surface leaning was correlated with strategic learning (-0.16) and attitude (-0.18). All these aforementioned correlations were in same direction in both sexes.
3) and 4) Linear regression model
model_1 <- lm(points ~ attitude + stra + surf, data = lrn14)
summary(model_1)##
## Call:
## lm(formula = points ~ attitude + stra + surf, data = lrn14)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.1550 -3.4346 0.5156 3.6401 10.8952
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.01711 3.68375 2.991 0.00322 **
## attitude 0.33952 0.05741 5.913 1.93e-08 ***
## stra 0.85313 0.54159 1.575 0.11716
## surf -0.58607 0.80138 -0.731 0.46563
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared: 0.2074, Adjusted R-squared: 0.1927
## F-statistic: 14.13 on 3 and 162 DF, p-value: 3.156e-08model_2 <- lm(points ~ attitude, data = lrn14)
summary(model_2)##
## Call:
## lm(formula = points ~ attitude, data = lrn14)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.9763 -3.2119 0.4339 4.1534 10.6645
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.63715 1.83035 6.358 1.95e-09 ***
## attitude 0.35255 0.05674 6.214 4.12e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared: 0.1906, Adjusted R-squared: 0.1856
## F-statistic: 38.61 on 1 and 164 DF, p-value: 4.119e-09Initially, I chose attitude, strategic learning and surface learning as explanatory variables for points in linear regression model as they were the variables showing highest correlation with points. However, attitude was the only statistically significant predictor of points (Coeff 0.34, p-value <0.001), while the model explained 20% of variation in points (Adjusted R squared). After removing non significant variables from the model, attitude predicted points with Coeff 0.35 (p-value <0.001), while the model explained 19% of variation in points. Taken together, based on this model, attitude was the best predictor of points: better attitude predicted more points.
5) Diagnostic plots
Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage
par(mfrow = c(2,2))
plot(model_2, which = c(1, 2, 5))
The main assumptions of Linear regression model are:
* linear relationship between response value and eplanatory value
* errors (residuals) have constant variance across all values of eplanatory variable
* errors are independent of each other
* errors have a normal distribution
Residuals vs fitted should not show a pattern where distribution of residuals varies along fitted values. In the figure above residuals are nicely evenly distributed.
QQ-plot should show a line if residuals are normally distributed. Figure above shows that the data is pretty much following the line excluding few outliers.
Residuals vs leverage should not show points outside Cook’s distance, which holds for the figure above.
Taken together, the assumptions of linear regression model hold true for the current model.
Chapter 3) Logistic regrsession
date()## [1] "Sun Nov 27 14:13:45 2022"Reading data
The data consists of student performance data in two Portuguese schools, collected by reports and questionnaires. It consists of two datasets from two different subjects: math and Portuguese. These datasets were combined and the variables of the combined dataset are printed below. In addition, two variables were calculated: alc_use is the average alcohol use from weekdays and weekends and high_use tells is true when alc_use is over 2
More information (including variable information) and the original dataset can be found from UCI_machine_learning_repository
alc <- read.csv("C:/Users/labpaavo/IODS-project/data/alc.csv")
colnames(alc)## [1] "school" "sex" "age" "address" "famsize"
## [6] "Pstatus" "Medu" "Fedu" "Mjob" "Fjob"
## [11] "reason" "guardian" "traveltime" "studytime" "schoolsup"
## [16] "famsup" "activities" "nursery" "higher" "internet"
## [21] "romantic" "famrel" "freetime" "goout" "Dalc"
## [26] "Walc" "health" "failures" "paid" "absences"
## [31] "G1" "G2" "G3" "alc_use" "high_use"Variables of interest in relation to alcohol consumption and hypotheses
For the variables of interest in relation to high/low alcohol consumption I have chosen: sex, Pstatus, studytime and absences. The hypotheses are that:
- being a male predicts high alcohol consumption
- parents living apart predicts high alcohol consumption
- low studytime predicts high alcohol consumption
- high absences predicts high alcohol consumption
Exploring the chosen variables
Numerically and graphically explore the distributions of your chosen variables and their relationships with alcohol consumption (use for example cross-tabulations, bar plots and box plots). Comment on your findings and compare the results of your exploration to your previously stated hypotheses. (0-5 points)
library(descr)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(ggplot2)
my_variables = c("high_use", "sex", "Pstatus", "studytime", "absences")
my_data <- alc[my_variables]
summary(my_data)## high_use sex Pstatus studytime
## Mode :logical Length:370 Length:370 Min. :1.000
## FALSE:259 Class :character Class :character 1st Qu.:1.000
## TRUE :111 Mode :character Mode :character Median :2.000
## Mean :2.043
## 3rd Qu.:2.000
## Max. :4.000
## absences
## Min. : 0.000
## 1st Qu.: 1.000
## Median : 3.000
## Mean : 4.511
## 3rd Qu.: 6.000
## Max. :45.000CrossTable(my_data$high_use, my_data$sex)## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
## =========================================
## my_data$sex
## my_data$high_use F M Total
## -----------------------------------------
## FALSE 154 105 259
## 2.244 2.500
## 0.595 0.405 0.700
## 0.79 0.60
## 0.416 0.284
## -----------------------------------------
## TRUE 41 70 111
## 5.235 5.833
## 0.369 0.631 0.300
## 0.21 0.40
## 0.111 0.189
## -----------------------------------------
## Total 195 175 370
## 0.527 0.473
## =========================================CrossTable(my_data$high_use, my_data$Pstatus)## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
## =========================================
## my_data$Pstatus
## my_data$high_use A T Total
## -----------------------------------------
## FALSE 26 233 259
## 0.014 0.002
## 0.100 0.900 0.700
## 0.684 0.702
## 0.070 0.630
## -----------------------------------------
## TRUE 12 99 111
## 0.032 0.004
## 0.108 0.892 0.300
## 0.316 0.298
## 0.032 0.268
## -----------------------------------------
## Total 38 332 370
## 0.103 0.897
## =========================================CrossTable(my_data$high_use, my_data$studytime)## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
## =========================================================
## my_data$studytime
## my_data$high_use 1 2 3 4 Total
## ---------------------------------------------------------
## FALSE 56 128 52 23 259
## 2.314 0.017 2.381 0.889
## 0.216 0.494 0.201 0.089 0.700
## 0.571 0.692 0.867 0.852
## 0.151 0.346 0.141 0.062
## ---------------------------------------------------------
## TRUE 42 57 8 4 111
## 5.400 0.041 5.556 2.075
## 0.378 0.514 0.072 0.036 0.300
## 0.429 0.308 0.133 0.148
## 0.114 0.154 0.022 0.011
## ---------------------------------------------------------
## Total 98 185 60 27 370
## 0.265 0.500 0.162 0.073
## =========================================================my_data %>% group_by(high_use) %>% summarise(count = n(), mean_absences = mean(absences))## # A tibble: 2 × 3
## high_use count mean_absences
## <lgl> <int> <dbl>
## 1 FALSE 259 3.71
## 2 TRUE 111 6.38g1 <- ggplot(data = my_data, aes(x = sex))
g1 + geom_bar() + facet_wrap("high_use") 
g2 <- ggplot(data = my_data, aes(x = Pstatus))
g2 + geom_bar() + facet_wrap("high_use") 
g3 <- ggplot(data = my_data, aes(x = studytime))
g3 + geom_bar() + facet_wrap("high_use") 
g4 <- ggplot(data = my_data, aes(x = absences))
g4 + geom_bar() + facet_wrap("high_use") 
Looking at data, it seems that first of all there are more students with low alcohol consumption (259) compared to high- (111). Absences is clearly skewed to left (less absences). It seems that high consumption group has more males (63%) compared to low consumption (41%). Parental status seems to be similarly distributed between the grpoups. In regards to studytime, high consumption group has notably higher frequency of students who study very little (37.8% vs 21.6%). Finally, high consumption group has higher mean absences (6.4), compared to low consumption group (3.7). Taken together, it seems that sex, studytime and absences could maybe predict alcohol consumptions, whereas parental status seems to be pretty evenly distributed between groups and thus will unlikely predict alcohol consumption.
Logistic regression
m <- glm(high_use ~ sex + Pstatus + studytime + absences, data = my_data, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ sex + Pstatus + studytime + absences,
## family = "binomial", data = my_data)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.1339 -0.8389 -0.5966 1.0728 2.1221
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.01835 0.55916 -1.821 0.068574 .
## sexM 0.84663 0.25509 3.319 0.000903 ***
## PstatusT 0.12379 0.40420 0.306 0.759415
## studytime -0.41526 0.16023 -2.592 0.009550 **
## absences 0.08928 0.02345 3.807 0.000141 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 408.40 on 365 degrees of freedom
## AIC: 418.4
##
## Number of Fisher Scoring iterations: 4OR <- coef(m) %>% exp
CI <- confint(m) %>% exp## Waiting for profiling to be done...cbind(OR, CI)## OR 2.5 % 97.5 %
## (Intercept) 0.3611909 0.1172496 1.0606103
## sexM 2.3317700 1.4202152 3.8687722
## PstatusT 1.1317735 0.5251790 2.5945605
## studytime 0.6601657 0.4776116 0.8967295
## absences 1.0933826 1.0465506 1.1475274Logistic regression models showed that sex, studytime and absences were significant predictors high/low alcohol consumptions:
sexM log odds .85 (p-value <0.001) studytime log odds -.42 (p-value < 0.01), meaning that for unit increase in studytime the log odds for being n high alcohol consumption decrease by 0.42.
absences log odds .09 (p-value <0.001), meaning that for each absence the log odds for high alcohol consumption changes by 0.09.
The odds for a male being in high alcohol consumption groups over a female is 2.3 (CI 1.4; 3.9). For unit increase in studytime the odds of being in high alcohol consumption group decrese by 34% (CI 0.48; 0.90), note that studytime had four possible values corresponging to <2 hours, 2 to 5 hours, 5 to 10 hours and >10 hours. For each absence the odds of being in high alcohol consumption group increase by 9.3% (CI 4.7%; 14.8%).
Against the initial hypothesis parental status did not predict alcohol consumption, however this was could be predicted already by exploring the data. The effect of sex, studytime and absences were as hypothesized as being a male, lower studytime and higher absences predicted being in high alcohol consumption group.
Predictive power of the model
# drop parental status from the model as it wasn't statistically significant predictor
new_model <- glm(high_use ~ sex + studytime + absences, data = my_data, family = "binomial")
probabilities <- predict(new_model, type = "response")
my_data <- mutate(my_data, probability = probabilities)
my_data <- mutate(my_data, prediction = (probability > 0.5))
table(high_use = my_data$high_use, prediction = my_data$prediction)## prediction
## high_use FALSE TRUE
## FALSE 250 9
## TRUE 86 25loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# training error
loss_func(class = my_data$high_use, prob = my_data$probability)## [1] 0.2567568g <- ggplot(my_data, aes(x = probability, y = high_use, col = prediction))
g + geom_point()
The model resulted in 250 true negatives, 25 true positives, 86 false negatives and 9 false positives.
The training error of the model is 25.7%.
10-fold cross-validation
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# compute the average number of wrong predictions in the (training) data
loss_func(class = my_data$high_use, prob = my_data$probability)## [1] 0.2567568# K-fold cross-validation
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.2648649The test set error of my model with 10-fold cross-validation is ~0.27, which is slightly worse than the model introduced in Exercise set.
Chapter 4) Clustering
date()## [1] "Sun Nov 27 14:13:48 2022"Reading data and exploring it
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectlibrary(corrplot)## corrplot 0.92 loadedlibrary(tidyr)
library(ggplot2)
data("Boston")
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14Boston data consists of 14 variables (and 506 observations) and it is about housing values in suburbs of Boston. Details and full descriptions of the variables can be found from Boston.
Graphical overview of the data
#summaries of the variable data
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00pairs(Boston)
#library(Hmisc)
boston_df <- as.data.frame(Boston)
#hist.data.frame(boston_df) this plot gives an error while knitting the index file so unfortunately can't display it
cor_matrix <- cor(Boston) %>% round(digits = 2)
corrplot(cor_matrix, method="circle", type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)
The figure of pairs function was so difficult to read on my laptop, so I ended up drawing also histograms to look at the distributions of the variables (unfortunately it can’t be displayed while knitting index because of the plot size, code commented out). Many variables are heavily skewed. The only variables close to normal distribution seem to be ‘average number of rooms per dwelling’(rm) and ‘median value of owner-occupied homes in $1000s’ (medv).
The correlation plot shows the relationships between the variables. Strongest negative correlations can be found between:
* weighted mean of distances to five Boston employment centres (dis) and proportion of owner-occupied units built prior to 1940 (age)
* dis and nitrogen oxides concentration (nox)
* dis and proportion of non-retail business acres per town (indus)
* lower status of the population percent (lastat) and median value of owner-occupied homes in $1000s (medv)
Strongest positive correlation is between index of accessibility to radial highways (rad) and full-value property-tax rate per $10,000 (tax)
Standardize the dataset
boston_scaled <- as.data.frame(scale(Boston))
boston_scaled$crim <- as.numeric(boston_scaled$crim)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
ind <- sample(nrow(boston_scaled), size = nrow(boston_scaled) * 0.8)
train <- boston_scaled[ind,]
test <- boston_scaled[-ind,]
correct_classes <- test$crime
test <- dplyr::select(test, -crime)
boston_scaled$crime <- factor(boston_scaled$crime, levels = c("low", "med_low", "med_high", "high"))Linear discriminant analysis
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "black", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes)
lda.arrows(lda.fit, myscale = 1)
Predict classes with LDA model
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class) ## predicted
## correct low med_low med_high high
## low 15 9 1 0
## med_low 4 18 4 0
## med_high 0 8 17 0
## high 0 0 1 25LDA model seems to predict classes quite well with accuracy of ~75% (varying between seeds). The model performed best in classifying high crime rates.
k-means clustering
# reloading the dataset
library(MASS)
data("Boston")
# standardizing the dataset
boston_scaled2 <- as.data.frame(scale(Boston))
# distances between observations
dist_eu <- dist(boston_scaled2)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970Looking for optimal number of clusters
library(ggplot2)
set.seed(13)
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled2, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
The optimal number of clusters is when the value of total WCSS (y-axis) changes radically, here it seems to be around two clusters.
# k-means clustering
km <- kmeans(boston_scaled2, centers = 2)
# plot the Boston dataset with clusters
pairs(boston_scaled2[1:6], col = km$cluster) 
pairs(boston_scaled2[7:14], col = km$cluster) 
I plotted the clusters in two separate plots, since the figures are otherwise two small for me to see in laptop, but this way I don’t see all the pairs. Overall it looks like two clusters works nicely in most of the pairs.
Super-Bonus:
matrix product
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layout# color is the crime classes of exercise set: classes is determined earlier as classes <- as.numeric(train$crime)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = classes)